Given: log627=a As we know that, logaxp=plogax ⇒3log63=a As we know that logab=
1
logba
⇒3∕log36=a⇒log36=3∕a As we know that, loga(x.y)=logax+logay ⇒log3(2×3)=log32+log33=3∕a As we know that, logaa=1 ⇒log32=3−a∕a ⇒log316=4×log32 Now by substituting the value of log32 in equation (1), we get ⇒log316=